commutes. Note that naturality of any such α\alpha implies that for all i,j∈Ji,j\in J, αi=αj\alpha_i=\alpha_j, so that α=Δ(ϕ)\alpha=\Delta(\phi) for some ϕ:c→c′\phi : c \to c' in CC. The single component ϕ\phi itself is often referred to as the cone morphism.

Discussion

Finn Lawler: I think the category of cones over FF is the comma category Δ/F\Delta / F, so that a morphism α:T→T′\alpha : T \to T' should be just a natural transformation α:Δc⇒Δc′\alpha : \Delta c \Rightarrow \Delta c' such that T′α=TT' \alpha = T. That gives your condition in components, I think.

Eric: Thanks Finn! I’m still learning all this, so it’ll take me some time to absorb what you said. It sounds good though :) Either way, it sounds like some potentially good additional content.

Finn Lawler: I should point out that a natural transformation α:Δc⇒Δc′\alpha: \Delta c \Rightarrow \Delta c' is very nearly exactly the same thing as a morphism ϕ:c→c′\phi: c \to c' (it’s ϕ\phi in each component, which you’ll see if you draw α\alpha‘s naturality square, so it’s Δϕ\Delta \phi for some ϕ\phi). Now look at the triangle above, write Δϕ:Δc→Δc′\Delta \phi : \Delta c \to \Delta c' instead of ϕ\phi and erase the jjs and you have the morphism in the comma category.

I hope this helps. If it’s done the opposite, apologies. I’ve a habit of trying the one and accomplishing the other.

Eric: Hmm. I’m probably confused, but when I draw the naturality square for α:Δ(c)→Δ(c′)\alpha:\Delta(c)\to\Delta(c'), I get

Eric: I think I got it. My diagram is correct, except we have αj=αk\alpha_j = \alpha_k and we want this to be ϕ:c→c′\phi:c\to c'. I made that more explicit in the definition above by adding “whose component is ϕ:c→c′\phi:c\to c'.”